Gabriel’s Horn

Here is one of my favorite “paradoxes” in mathematics, that many people will learn in a first year calculus course. It is called “Gabriel’s Horn.”

First, we take the function f(x) = \dfrac{1}{x}. This is a curve that we can imagine beginning at the point (1,1), then quickly sloping down towards the x-axis at which point it becomes nearly horizontal. A plot is shown below.

prerotation

Now imagine taking that curve and rotating it around the x-axis, forming an infinitely long shape that looks like the bell of a trumpet, or horn. After doing so, we get a shape that looks roughly like this:

geogebra-export(1)

There are two things we wish to determine about this new three-dimensional figure: its volume, and its surface area. We can think of the volume simply as how much stuff we need to fill it completely, while the surface area is how much paint we need to coat the surface. As we will show, this figure has finite volume but infinite surface area. What this means is that we would need an infinite amount of paint in order to just cover the inside surface with a layer of paint. However, if we really wanted to paint the inside, we could take our finite amount of paint and just “pour it in” the horn. We could fill up this infinite horn with a finite amount of paint, and thus end up painting the inside.

The reason this becomes fun is because this is an infinite object. The skinny part of the bell goes on forever, toward +\infty. But this is no obstacle for some tools in calculus. The main idea of calculus is to split your object into arbitrarily thin or short pieces. In our case, we will take an arbitrary vertical slice out of our horn at some horizontal point we’ll call x_1. The disk we get as a result will be circular, and have some radius r. Consider the figure below.

finalHorn

The radius is the distance from the x-axis to the curve f(x) = \dfrac{1}{x}. But by definition, this is just the y-coordinate, f(x_1) = \dfrac{1}{x_1}. So, for any arbitrary vertical slice taken out of our horn, we can find the radius. This allows us to easily find the circumference (2\pi r) and area (\pi r^2) of each disk.

At this point, we’ll have to assume some knowledge of calculus. Consider taking every possible arbitrarily thin disk in the horn, infinitely many of them! If we add their areas over and over again, \pi r^2 = \pi \left(\dfrac{1}{x}\right)^2, multiplied by their tiny piece of width we call dx, we will obtain the volume of the horn overall. We can do similarly with the circumference, multiplying each tiny bit of circumference, 2 \pi r = 2\pi \dfrac{1}{x}, with that same bit of width dx. This will give us our surface area.

We do this formally by using integration. In particular, we use an “improper” integral, and technically have to take a limit (of course, once I finished calculus I got too lazy to write out those steps normally, but I will now for completeness.)

For volume, our integral is

\displaystyle \text{Volume} =  \pi \int\limits_1^\infty \dfrac{1}{x^2}dx = \pi \lim\limits_{b\to\infty} \int\limits_1^b \dfrac{1}{x^2}dx

We can evaluate this integral directly, getting

\displaystyle \text{Volume} = \pi \lim\limits_{b\to\infty}\left. -\dfrac{1}{x} \right|_1^b = \pi \lim\limits_{b\to\infty} \left( -\dfrac{1}{b} + 1\right) = \boxed{\pi}.

That -\dfrac{1}{b} goes to 0, so that term disappears. As we stated earlier, the volume is finite. Choose your units, and the volume of that infinite horn above is just \pi. Neat!

We can do similarly for the surface area, but our integral will diverge.

\displaystyle \text{Surface Area} = 2\pi\int\limits_1^\infty \dfrac{1}{x}dx = 2\pi \lim\limits_{b\to\infty} \int_1^b \dfrac{1}{x}dx = 2\pi \lim\limits_{b\to\infty} \left(\ln b - \ln 1\right) = \boxed{\infty}.

Since \ln 1= 0, we just get \ln b going off to infinity, hence the surface area is infinite.

There are many cool problems like this, some with significantly less background necessary to formulate. I’m hoping to mix these in with some other types of posts.

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